Question #141256
If 50.9 g of aspirin (C₉H₈O₄) are produced from 79.8 g of C₇H₆O₃, what is the percent yield from the reaction below?
C₇H₆O₃ (s) + C₄H₆O₃ (s) → C₉H₈O₄ (s) + HC₂H₃O₂ (aq).
1
Expert's answer
2020-10-30T06:45:16-0400

M(C₉H₈O₄) = 180.15 g/mol

M(C₇H₆O₃) = 138.12 g/mol

n(C7H6O3)=79.8138.12=0.577  moln(C₇H₆O₃) = \frac{79.8}{138.12} = 0.577 \;mol

n(C7H6O3)=n(C9H8O4)=0.577  moln(C₇H₆O₃) = n(C₉H₈O₄) = 0.577 \;mol

m(C9H8O4)=0.577×180.15=103.94  gm(C₉H₈O₄) = 0.577 \times 180.15 = 103.94 \;g

Proportion:

103.94 – 100 %

50.9 – x

x=50.9×100103.94=48.97x = \frac{50.9 \times 100}{103.94} = 48.97 % (the percent yield from the reaction)

Answer: 48.97 %.

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